Skip to main content
Workforce LibreTexts

3.3: Solving Quadratic Equations of the Form- (x-a)²=c

  • Page ID
    20037
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    When solving equations of the form: \({\left(x-a\right)}^2=c\), we can use the square root property which states to drop the square and take the positive and negative square root of the constant, c. Mathematically, we say the following:

    If \({\left(x-a\right)}^2=c\),

    then \(x-a=\pm \sqrt{c}\)

    next, add a on both sides of the equation to obtain

    \(x=a\pm \sqrt{c}\)

    Example \(\PageIndex{1}\)

    Solve \({\left(x-5\right)}^2=16\)

    Solution

    To solve, we first drop the square from the left side and take the positive and negative square root of 16 to obtain

    \[x-5=\pm \sqrt{16}\]

    Next, we simplify the square root

    \[x-5=\pm 4\]

    Next, we isolate the variable, x by adding 5 on both sides of the equation:

    \[x-5+\boldsymbol{5}=+\boldsymbol{5}\pm 4\]

    Such that after simplifying, we obtain two answers

    \[x=5\pm 4\]

    \[x=5+4 \,\,\text{ and }\,\, x=5-4\]

    \[x=9\,\, \text{ and }\,\, x=1\]

    Example \(\PageIndex{1}\)

    Solve \(3{\left(x+2\right)}^2+5=80\)

    Solution

    To solve, we must first isolate the perfect square to create the form: \({\left(x-a\right)}^2=c\), so we first subtract 5 on both sides of the equation, then divide both sides by 3

    \[3{\left(x+2\right)}^2+5-\boldsymbol{5}=80-\boldsymbol{5}\]

    \[3{\left(x+2\right)}^2=75\]

    \[\frac{3\left(x+2\right)^2}{\boldsymbol{3}}=\frac{75}{\boldsymbol{3}}\]

    \[{\left(x+2\right)}^2=25\]

    Next, we drop the square and take the positive and negative square root of 25 to obtain

    \[x+2=\pm \sqrt{25}\]

    \[x+2=\pm 5\]

    Next, we isolate the x by subtracting 2 on both sides of the equation

    \[x+2-\boldsymbol{2}=-\boldsymbol{2}\pm 5\]

    After simplifying, we obtain two answers

    \[x=-2\pm 5\]

    \[x=-2+5 \,\,\text{ and }\,\, x=-2-5\]

    \[x=3 \,\,\text{ and }\,\, x=-7\]


    This page titled 3.3: Solving Quadratic Equations of the Form- (x-a)²=c is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Kelly Brooks.